mapping (0,1) onto [0,1]: any such continuous function?
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skeeter
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Re: mapping (0,1) onto [0,1]
only one I can think of off the top of my head would have to be defined piece-wise ... is that satisfactory?
\(\displaystyle f(x) = 1\)
\(\displaystyle 0 < x \leq \frac{1}{4}\)
\(\displaystyle f(x) = -2x + \frac{3}{2}\)
\(\displaystyle \frac{1}{4} < x < \frac{3}{4}\)
\(\displaystyle f(x) = 0\)
\(\displaystyle \frac{3}{4} \leq x < 1\)
only one I can think of off the top of my head would have to be defined piece-wise ... is that satisfactory?
\(\displaystyle f(x) = 1\)
\(\displaystyle 0 < x \leq \frac{1}{4}\)
\(\displaystyle f(x) = -2x + \frac{3}{2}\)
\(\displaystyle \frac{1}{4} < x < \frac{3}{4}\)
\(\displaystyle f(x) = 0\)
\(\displaystyle \frac{3}{4} \leq x < 1\)